AP Chemistry 9.5-9.6: Free Energy and Equilibrium, Coupled Reactions

Gibbs free energy

A thermodynamic state function defined by $G = H - TS$. At constant temperature and pressure, its change is $ΔG = ΔH - TΔS$.

Why is Gibbs free energy useful for predicting spontaneity?

It determines spontaneity using properties of the system alone, without separately measuring the entropy change of the surroundings. At constant temperature and pressure, $ΔG$ is related to the entropy change of the universe by $ΔG = -TΔS_{univ}$.

How do the signs of $ΔG$ and $ΔS_{univ}$ indicate spontaneity at constant temperature and pressure?

If $ΔG < 0$ and $ΔS_{univ} > 0$, the process is spontaneous in the forward direction. If $ΔG > 0$ and $ΔS_{univ} < 0$, it is nonspontaneous as written; if $ΔG = 0$, the system is at equilibrium.

What does a positive or negative $ΔG$ imply about the favored direction of a process?

A negative $ΔG$ provides a thermodynamic driving force in the forward direction, whereas a positive $ΔG$ favors the reverse direction. This prediction concerns thermodynamic favorability, not the reaction rate.

Standard free energy change, $ΔG^\circ$

The Gibbs free energy change for a process when all substances are in their standard states, commonly at a specified temperature such as $298\ \mathrm{K}$. Its sign predicts spontaneity only under those standard-state conditions.

How can standard free energy change be calculated from standard enthalpy and entropy changes?

Use $ΔG^\circ = ΔH^\circ - TΔS^\circ$, with $T$ in kelvins. Energy units must be consistent, so entropy in $\mathrm{J\,mol^{-1}\,K^{-1}}$ is often converted to $\mathrm{kJ\,mol^{-1}\,K^{-1}}$.

Standard Gibbs free energy of formation, $ΔG_f^\circ$

The free energy change for forming one mole of a substance from its elements in their standard states. For an element in its standard state, $ΔG_f^\circ = 0$.

How is a reaction's $ΔG^\circ$ calculated from standard Gibbs free energies of formation?

Use $ΔG^\circ = \sum \nuΔG_f^\circ(\mathrm{products}) - \sum \nuΔG_f^\circ(\mathrm{reactants})$, where each formation free energy is multiplied by its stoichiometric coefficient.

Why can reaction free energies be calculated by adding formation reactions?

Gibbs free energy is a state function, so its change depends only on the initial and final states. Therefore, free energy changes add according to Hess's law when reactions are combined, reversed, or multiplied.

How should the free energy change of a reversed reaction be related to the original reaction?

Reversing a reaction changes the sign of its free energy change: $ΔG_{reverse} = -ΔG_{forward}$. Multiplying a reaction by a factor multiplies $ΔG$ by the same factor.

Reaction coupling

Combining a thermodynamically unfavorable reaction with a favorable reaction so that the sum has a negative overall $ΔG$. The free energy changes of the component reactions are added after canceling intermediates.

How can a nonspontaneous reaction become favorable through coupling?

If an unfavorable reaction has positive $ΔG$ and a coupled reaction has a sufficiently negative $ΔG$, the sum can be spontaneous. For example, a reaction with $ΔG_1 = +201.3\ \mathrm{kJ}$ coupled to one with $ΔG_2 = -300.1\ \mathrm{kJ}$ gives $ΔG_{overall} = -98.8\ \mathrm{kJ}$.

How do the signs of $ΔH$ and $ΔS$ predict the temperature dependence of spontaneity?

For $ΔH<0$ and $ΔS>0$, the process is spontaneous at all temperatures. For $ΔH>0$ and $ΔS<0$, it is nonspontaneous at all temperatures. If both have the same sign, spontaneity depends on temperature.

For a process with $ΔH>0$ and $ΔS>0$, when is it spontaneous?

Because $ΔG = ΔH-TΔS$, increasing temperature makes the negative $-TΔS$ term more important. The process is spontaneous at sufficiently high temperatures and nonspontaneous at sufficiently low temperatures.

For a process with $ΔH<0$ and $ΔS<0$, when is it spontaneous?

The process is spontaneous at sufficiently low temperatures because the negative enthalpy term dominates. At sufficiently high temperatures, the positive contribution from $-TΔS$ can make $ΔG$ positive.

At what temperature is a process with temperature-dependent spontaneity at equilibrium?

At the transition temperature, $ΔG=0$, so $0=ΔH-TΔS$ and $T=\frac{ΔH}{ΔS}$. Use consistent energy units for $ΔH$ and $ΔS$, and use kelvins for $T$.

Why does the equation $T=\frac{ΔH}{ΔS}$ estimate a phase-transition temperature?

At the equilibrium temperature for a phase transition, the two phases have equal free energy, so $ΔG=0$. Substitution into $ΔG=ΔH-TΔS$ gives $T=\frac{ΔH}{ΔS}$.

How is free energy under nonstandard conditions related to standard free energy?

Use $ΔG = ΔG^\circ + RT\ln Q$, where $Q$ is the reaction quotient and $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$. This equation predicts spontaneity for the actual composition, pressure, and temperature.

Reaction quotient and free energy

The reaction quotient $Q$ is calculated using the same mass-action expression as $K$, but with the current, nonequilibrium quantities. Its value determines how the current composition shifts the free energy relative to $ΔG^\circ$.

Which reaction quotient should be used for gas-phase and condensed-phase equilibria?

Use the pressure-based quotient $Q_P$ for gas-phase equilibria and the concentration-based quotient $Q_C$ for equilibria involving dissolved species. Pure solids and liquids are omitted from the quotient.

How can $Q$ and $K$ be used to predict the direction of a reaction?

If $Q<K$, the reaction proceeds forward until $Q=K$. If $Q>K$, it proceeds in reverse; if $Q=K$, the system is at equilibrium.

What is the relationship between $ΔG$, $Q$, and the direction of spontaneous change?

For the current conditions, $ΔG<0$ means the forward reaction is spontaneous, while $ΔG>0$ means the reverse reaction is spontaneous. At equilibrium, $Q=K$ and $ΔG=0$.

How are standard free energy change and the equilibrium constant related?

At equilibrium, $ΔG=0$ and $Q=K$, giving $ΔG^\circ=-RT\ln K$. Equivalently, $K=e^{-ΔG^\circ/(RT)}$.

What does the sign and magnitude of $ΔG^\circ$ imply about $K$?

If $ΔG^\circ<0$, then $K>1$ and products are favored at equilibrium. If $ΔG^\circ>0$, then $K<1$ and reactants are favored; if $ΔG^\circ=0$, then $K=1$.

Why does a negative $ΔG^\circ$ not mean that every mixture spontaneously forms only products?

A negative $ΔG^\circ$ describes the reaction under standard conditions and indicates $K>1$, not complete conversion. The actual direction under any composition depends on $ΔG=ΔG^\circ+RT\ln Q$.

How does a system reach equilibrium from a nonequilibrium composition?

The reaction proceeds spontaneously in whichever direction lowers the system's Gibbs free energy. Equilibrium is reached when free energy is minimized and the forward and reverse driving forces are equal, so $Q=K$ and $ΔG=0$.